There are a couple different ways to combine random variables into a new random variable: **means** and **mixtures**. To take the mean of *X* and *Y* you average their **values**. To take the mixture of *X* and *Y* you average their **densities**. The former makes the tails thinner. The latter makes the tails thicker. When *X* and *Y* are exponential random variables, the mean has a **hypoexponential** distribution and the mixture has a **hyperexponential** distribution.

## Hypoexponential distributions

Suppose *X* and *Y* are exponentially distributed with mean μ. Then their sum *X* + *Y* has a gamma distribution with shape 2 and scale μ. The sum has mean 2μ and variance 2μ². The **coefficient of variation**, the ratio of the standard deviation to the mean, is 1/√2. The hypoexponential distribution is so-called because its coefficient of variation is less than 1, whereas an exponential distribution has coefficient of variation 1 because the mean and standard deviation are always the same.

The means of *X* and *Y* don’t have to be the same. When they’re different, the sum does not have a gamma distribution, and so hypoexponential distributions are more general than gamma distributions.

A hypoexponential random variable can also be the sum of more than two exponentials. If it is the sum of *n* exponentials, each with the same mean, then the coefficient of variation is 1/√*n*. In general, the coefficient of variation for a hypoexponential distribution is the coefficient of variation of the means [1].

In the introduction we talked about means rather than sums, but it makes no difference to the coefficient of variation because going from sum to mean divides the mean and the standard deviation by the same amount.

## Hyperexponential distributions

Hyperexponential random variables are constructed as a mixture of exponentials rather than an average. Instead selecting a value from *X* **and** a value from *Y*, we select a value from *X* **or** a value from *Y*. Given a mixture probability *p*, we choose a sample from *X* with probability *p* and a value from *Y* with probability 1 – *p*.

The density function for a mixture is a an average of the densities of the two components. So if *X* and *Y* have density functions *f*_{X} and *f*_{Y}, then the mixture has density

*p* *f*_{X} + (1 – *p*) *f*_{Y}

If you have more than two random variables, the distribution of their mixture is a convex combination of their individual densities. The coefficients in the convex combination are the probabilities of selecting each random variable.

If *X* and *Y* are exponential with means μ_{X} and μ_{Y}, and we have a mixture that selects *X* with probability *p*, then then mean of the mixture is the mixture of the means

μ = *p* μ_{X} + (1 – *p*) μ_{Y}

which you might expect, but the variance

σ² = *p* μ_{X} ² + (1 – *p*) μ_{Y} ² + *p*(1 – *p*)(μ_{X} – μ_{Y})²

is not quite analogous because of the extra *p*(1 – *p*)(μ_{X} – μ_{Y})² term at the end. If μ_{X} = μ_{Y} this last term drops out and the coefficient of variation is 1: mixing two identically distributed random variables doesn’t do anything to the distribution. But when the means are different, the coefficient of variation is greater than 1 because of the extra term in the variance of the mixture.

## Example

Suppose μ_{X} = 2 and μ_{Y} = 8. Then the average of *X* and Y has mean 5, and so does an equal mixture of *X* and *Y*.

The average of *X* and *Y* has standard deviation √17, and so the coefficient of variation is √17/5 = 0.8246.

An exponential distribution with mean 5 would have standard deviation 5, and so the coefficient of variation 1.

An equal mixture of *X* and *Y* has standard deviation √43, and so the coefficient of variation is √43/5 = 1.3114.

## Related posts

[1] If exponential random variables *X*_{i} have means μ_{i}, then the coefficient of variation of their sum (or average) is

√(μ_{1}² + μ_{2}² + … + μ_{n}²) / (μ_{1} + μ_{2} + … + μ_{n})